IS THERE A FENCEPOST ERROR WHEN CALCULATING MASS FOR A CONTINUOUS DISTRIBUTION?

[LaplacianHarmonic]

Let's say 5 masses are arranged on x axis.

At x=1, 2kg
X=2, 4kg
X=3, 6kg
X=4, 8kg
X=5, 10 kg

Obviously, there is a total mass of 30 kgIf the mass is distributed continuously by the function M(x) = 2x, then

From x= 0 to x=5, there is 25 kg of mass from the simple integral.

WHY IS THERE LESS MASS for the continuous mass distribution than the 5 pint masses that equal up to 30kg?

is there NOT more mass for the continuous mass distribution than the 5 finite point masses? (In between the integers along the x axis?)

 

[mathman]

The point masses are at the upper limits of the intervals and have their values at the upper limit values. The continuous distribution for each interval, when integrated, is less than the upper limit value.

 

[jbriggs444]

WHY IS THERE LESS MASS for the continuous mass distribution than the 5 pint masses that equal up to 30kg?

One way of looking at it is as a "fencepost error".

Think of spreading the mass at each integer out. So that the 2 kg at x=1 is spread across from x=0.5 to x=1.5 and the 10 kg at x=5 is spread across from x=4.5 to x=5.5. If you wish, consider 0 kg at x=0 spread across from x=-0.5 to x=+0.5

Now instead of a distribution with a series of spikes you have a distribution that is a step function. The total mass is still the same.

Now change to a continuous distribution, smoothing out that step function. Again, the total mass is the same.

But now look at the integrals that you have used to try to compute the total mass. You summed from 0 to 5 and counted the endpoints at full value. You integrated from 0 to 5. But that step function goes from -0.5 to +5.5. You've chopped off half of the first and last interval.

Try doing the sum by counting the first and last masses at half-value: $\frac{0}{2}$ + 2 + 4 + 6 + 8 + $\frac{10}{2}$ Now you get 25, just like the integral.
Or integrate from -0.5 to +5.5. Now you get 30, just like the sum.